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# characterization of field

###### Proposition 1.

Let $\mathcal{R}\neq 0$ be a commutative ring with identity. The ring $\mathcal{R}$ (as above) is a field if and only if $\mathcal{R}$ has exactly two ideals: $(0),\mathcal{R}$.

###### Proof.

($\Rightarrow$) Suppose $\mathcal{R}$ is a field and let $\mathcal{A}$ be a non-zero ideal of $\mathcal{R}$. Then there exists $r\in\mathcal{A}\subseteq\mathcal{R}$ with $r\neq 0$. Since $\mathcal{R}$ is a field and $r$ is a non-zero element, there exists $s\in\mathcal{R}$ such that

$s\cdot r=1\in\mathcal{R}$ |

Moreover, $\mathcal{A}$ is an ideal, $r\in\mathcal{A},s\in\mathcal{S}$, so $s\cdot r=1\in\mathcal{A}$. Hence $\mathcal{A}=\mathcal{R}$. We have proved that the only ideals of $\mathcal{R}$ are $(0)$ and $\mathcal{R}$ as desired.

($\Leftarrow$) Suppose the ring $\mathcal{R}$ has only two ideals, namely $(0),\mathcal{R}$. Let $a\in\mathcal{R}$ be a non-zero element; we would like to prove the existence of a multiplicative inverse for $a$ in $\mathcal{R}$. Define the following set:

$\mathcal{A}=(a)=\{r\in\mathcal{R}\mid r=s\cdot a,\text{ for some }s\in\mathcal{R}\}$ |

This is clearly an ideal, the ideal generated by the element $a$. Moreover, this ideal is not the zero ideal because $a\in\mathcal{A}$ and $a$ was assumed to be non-zero. Thus, since there are only two ideals, we conclude $\mathcal{A}=\mathcal{R}$. Therefore $1\in\mathcal{A}=\mathcal{R}$ so there exists an element $s\in\mathcal{R}$ such that

$s\cdot a=1\in\mathcal{R}$ |

Hence for all non-zero $a\in\mathcal{R}$, $a$ has a multiplicative inverse in $\mathcal{R}$, so $\mathcal{R}$ is, in fact, a field. ∎

## Mathematics Subject Classification

12E99*no label found*

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## Corrections

Proof included by Larry Hammick ✓

assumption by matte ✓

could be more general by scineram ✘